if dydx cosx cos2 y and ypi4 when x0 thenSolutiondydx cosx

Solution

dy/dx = cos(x) cos^2(y)

1/cos^2(y) dy = cos(x) dx

(1+tan^2(y)) dy = cos(x) dx

int 1+tan^2(y) dy = int cos(x) dx

tan(y) = sin(x) + C

x = 0 -> tan(pi/4) = sin(0) + C -> 1 = 0+C -> C = 1

So the solution is:

tan(y) = sin(x)+1

or

y = arctan(sin(x)+1)